Question

A liquid of mass 100 g at 120 °c is poured in water at 20°c, when the final temp recorded is 40°c. If the specific heat capacity of liquid is 0.8 J/g/°c, calculate the initial mass of water?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Heat}\mathrm{given}\mathrm{out}\mathrm{by}\mathrm{liquid}={\mathrm{mC\theta }}_{\mathrm{f}} \\ \mathrm{Heat}\mathrm{given}\mathrm{out}\mathrm{by}\mathrm{liquid}=100\times 0.8\times \left(120-40\right)=6400\mathrm{J} \\ \mathrm{Heat}\mathrm{absorbed}\mathrm{by}\mathrm{water}={\mathrm{MC\theta }}_{\mathrm{R}} \\ \mathrm{Heat}\mathrm{absorbed}\mathrm{by}\mathrm{water}=\mathrm{M}\times 0.8\times \left(40-20\right) \\ \mathrm{Heat}\mathrm{given}\mathrm{out}\mathrm{by}\mathrm{liquid}=\mathrm{Heat}\mathrm{absorbed}\mathrm{by}\mathrm{water} \\ 6400=\mathrm{M}\times 0.8\times \left(40-20\right) \\ \mathrm{M}=\frac{6400}{0.8\times 20}=400\mathrm{g} \end{gathered}$$